Difference between revisions of "1955 AHSME Problems"

Revision as of 08:43, 13 October 2014

== Problem 1==

Which one of the following is not equivalent to $0.000000375$ ?

$\textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\\ \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000}$

Solution

Problem 2

The smaller angle between the hands of a clock at $12:25$ p.m. is:

$\textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ$

Solution

Problem 3

If each number in a set of ten numbers is increased by $20$ , the arithmetic mean (average) of the original ten numbers:

$\textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2}$

Solution

Problem 4

The equality $\frac{1}{x-1}=\frac{2}{x-2}$ is satisfied by:

$\textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0$

Solution

Problem 5

$5y$ varies inversely as the square of $x$ . When $y=16, x=1$ . When $x=8, y$ equals:

\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024

Solution

Problem 6

A merchant buys a number of oranges at $3$ for $10$ cents and an equal number at $5$ for $20$ cents. To "break even" he must sell all at:

$\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}$

Solution

Problem 7

If a worker receives a $20$ % cut in wages, he may regain his original pay exactly by obtaining a raise of:

$\textbf{(A)}\ \text{20%}\qquad\textbf{(B)}\ \text{25%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{%}\qquad\textbf{(D)}\ $20\qquad\textbf{(E)}\ $25$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 8

The graph of $x^2-4y^2=0$ :

$\textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does not exist}$

Solution

Problem 9

A circle is inscribed in a triangle with sides $8, 15$ , and $17$ . The radius of the circle is:

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7$

Solution

Problem 10

How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each?

$\textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40}$

Solution

Problem 11

The negation of the statement "No slow learners attend this school" is:

$\textbf{(A)}\ \text{All slow learners attend this school}\\ \textbf{(B)}\ \text{All slow learners do not attend this school}\\ \textbf{(C)}\ \text{Some slow learners attend this school}\\ \textbf{(D)}\ \text{Some slow learners do not attend this school}\\ \textbf{(E)}\ \text{No slow learners do not attend this school}$

Solution

Problem 12

The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:

$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$

Solution

Problem 13

The fraction $\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}$ is equal to:

$\textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2$

Solution

Problem 14

The length of rectangle $R$ is $10$ % more than the side of square $S$ . The width of the rectangle is $10$ % less than the side of the square. The ratio of the areas, $R:S$ , is:

$\textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200$

Solution

Problem 15

The ratio of the areas of two concentric circles is $1: 3$ . If the radius of the smaller is $r$ , then the difference between the radii is best approximated by:

$\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r$

Solution

Problem 16

The value of $\frac{3}{a+b}$ when $a=4$ and $b=-4$ is:

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless}$

Solution

Problem 17

If $\log x-5 \log 3=-2$ , then $x$ equals:

$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$

Solution

Problem 18

The discriminant of the equation $x^2+2x\sqrt{3}+3=0$ is zero. Hence, its roots are:

$\textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary}$

Solution

Problem 19

Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:

$\textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0$

Solution

Problem 20

The expression $\sqrt{25-t^2}+5$ equals zero for:

$\textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5$

Solution

Problem 21

Represent the hypotenuse of a right triangle by $c$ and the area by $A$ . The altitude on the hypotenuse is:

$\textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2}$

Solution

Problem 22

On a \texdollar{10000} order a merchant has a choice between three successive discounts of $20$ %, $20$ %, and $10$ % and three successive discounts of $40$ %, $5$ %, and $5$ %. By choosing the better offer, he can save:

$\textbf{(A)}\ \text{nothing at all}\qquad\textbf{(B)}\ $440\qquad\textbf{(C)}\ $330\qquad\textbf{(D)}\ $345\qquad\textbf{(E)}\ $360$

Solution

Problem 23

In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:

$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents}$

Solution

Problem 24

The function $4x^2-12x-1$ :

$\textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10}$

Solution

Problem 25

One of the factors of $x^4+2x^2+9$ is:

$\textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 26

Mr. A owns a house worth \textdollar{10000}. He sells it to Mr. $B$ at $10$ % profit. Mr. $B$ sells the house back to Mr. $A$ at a $10$ % loss. Then:

$\textbf{(A)}\ \text{Mr. A comes out even}\qquad\textbf{(B)}\ \text{Mr. A makes }$100\qquad\textbf{(C)}\ \text{Mr. A makes }$1000\\ \textbf{(D)}\ \text{Mr. B loses }$100\qquad\textbf{(E)}\ \text{none of the above is correct}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 27

If $r$ and $s$ are the roots of $x^2-px+q=0$ , then $r^2+s^2$ equals:

$\textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2$

Solution

Problem 28

On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation. If $b \neq 0$ and $c \neq 0$ these two graphs intersect:

$\textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis}$

Solution

Problem 29

In the figure, $PA$ is tangent to semicircle $SAR$ ; $PB$ is tangent to semicircle $RBT$ ; $SRT$ is a straight line; the arcs are indicated in the figure. $\angle APB$ is measured by:

$[asy] unitsize(1.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0); pair A=O1+2*dir(60), B=O2+dir(85); pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2; pair P=extension(A,Pa,B,Pb); pair[] dots={Sp,R,T,A,B,P}; draw(P--P+5*(A-P)); draw(P--P+5*(B-P)); clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle); draw(Arc(O1,2,0,180)--cycle); draw(Arc(O2,1,0,180)--cycle); dot(dots); label("$S$",Sp,S); label("$R$",R,S); label("$T$",T,S); label("$A$",A,NE); label("$B$",B,N); label("$P$",P,NNE); label("$a$",midpoint(Arc(O1,2,0,60)),SW); label("$b$",midpoint(Arc(O2,1,85,180)),SE); label("$c$",midpoint(Arc(O1,2,60,180)),SE); label("$d$",midpoint(Arc(O2,1,0,85)),SW);[/asy]$

$\textbf{(A)}\ \frac{1}{2}(a-b)\qquad\textbf{(B)}\ \frac{1}{2}(a+b)\qquad\textbf{(C)}\ (c-a)-(d-b)\qquad\textbf{(D)}\ a-b\qquad\textbf{(E)}\ a+b$

Solution

Problem 30

Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:

$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$

Solution

Problem 31

An equilateral triangle whose side is $2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:

$\textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2}$

Solution

Problem 32

If the discriminant of $ax^2+2bx+c=0$ is zero, then another true statement about $a, b$ , and $c$ is that:

$\textbf{(A)}\ \text{they form an arithmetic progression}\\ \textbf{(B)}\ \text{they form a geometric progression}\\ \textbf{(C)}\ \text{they are unequal}\\ \textbf{(D)}\ \text{they are all negative numbers}\\ \textbf{(E)}\ \text{only b is negative and a and c are positive}$

Solution

Problem 33

Henry starts a trip when the hands of the clock are together between $8$ a.m. and $9$ a.m. He arrives at his destination between $2$ p.m. and $3$ p.m. when the hands of the clock are exactly $180^\circ$ apart. The trip takes:

$\textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. 43-7/11 min.}\qquad\textbf{(C)}\ \text{5 hr. 16-4/11 min.}\qquad\textbf{(D)}\ \text{6 hr. 30 min.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 34

A $6$ -inch and $18$ -inch diameter pole are placed together and bound together with wire. The length of the shortest wire that will go around them is:

$\textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi$

Solution

Problem 35

Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:

$\textbf{(A)}\ \text{is none of the following}\qquad\textbf{(B)}\ \text{cannot be determined from the given data}\\ \textbf{(C)}\ \text{is 20 or 26}\qquad\textbf{(D)}\ \text{is 14 or 32}\qquad\textbf{(E)}\ \text{is 8 or 38}$

Solution

Problem 36

A cylindrical oil tank, lying horizontally, has an interior length of $10$ feet and an interior diameter of $6$ feet. If the rectangular surface of the oil has an area of $40$ square feet, the depth of the oil is:

$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 3-\sqrt{5}\qquad\textbf{(D)}\ 3+\sqrt{5}\\ \textbf{(E)}\ \text{either }3-\sqrt{5}\text{ or }3+\sqrt{5}$

Solution

Problem 37

A three-digit number has, from left to right, the digits $h, t$ , and $u$ , with $h>u$ . When the number with the digits reversed is subtracted from the original number, the units' digit in the difference of r. The next two digits, from right to left, are:

$\textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\qquad\textbf{(C)}\ \text{impossible to tell}\qquad\textbf{(D)}\ \text{5 and 4}\qquad\textbf{(E)}\ \text{4 and 5}$

Solution

Problem 38

Four positive integers are given. Select any three of these integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers $29, 23, 21$ , and $17$ are obtained. One of the original integers is:

\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17

Solution

Problem 39

If $y=x^2+px+q$ , then if the least possible value of $y$ is zero $q$ is equal to:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q$

Solution

Problem 40

The fractions $\frac{ax+b}{cx+d}$ and $\frac{b}{d}$ are unequal if:

$\textbf{(A)}\ a=c=1, x\neq 0\qquad\textbf{(B)}\ a=b=0\qquad\textbf{(C)}\ a=c=0\\ \textbf{(D)}\ x=0\qquad\textbf{(E)}\ ad=bc$

Solution

Problem 41

A train traveling from Aytown to Beetown meets with an accident after $1$ hr. It is stopped for $\frac{1}{2}$ hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown $2$ hr. late. If the train had covered $80$ miles more before the accident, it would have been just $1$ hr. late. The usual rate of the train is:

$\textbf{(A)}\ \text{20 mph}\qquad\textbf{(B)}\ \text{30 mph}\qquad\textbf{(C)}\ \text{40 mph}\qquad\textbf{(D)}\ \text{50 mph}\qquad\textbf{(E)}\ \text{60 mph}$

Solution

Problem 42

If $a, b$ , and $c$ are positive integers, the radicals $\sqrt{a+\frac{b}{c}}$ and $a\sqrt{\frac{b}{c}}$ are equal when and only when:

$\textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{2}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text{ and }c=a-1$

Solution

Problem 43

The pairs of values of $x$ and $y$ that are the common solutions of the equations $y=(x+1)^2$ and $xy+y=1$ are:

$\textbf{(A)}\ \text{3 real pairs}\qquad\textbf{(B)}\ \text{4 real pairs}\qquad\textbf{(C)}\ \text{4 imaginary pairs}\\ \textbf{(D)}\ \text{2 real and 2 imaginary pairs}\qquad\textbf{(E)}\ \text{1 real and 2 imaginary pairs}$

Solution

Problem 44

In circle $O$ chord $AB$ is produced so that $BC$ equals a radius of the circle. $CO$ is drawn and extended to $D$ . $AO$ is drawn. Which of the following expresses the relationship between $x$ and $y$ ?

$[asy] size(200);defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0); draw(O--A--C--D); dot(A^^B^^C^^D^^O); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(285)); label("$x$", O+0.1*dir(172.5), dir(172.5)); label("$y$", C+0.4*dir(187.5), dir(187.5)); draw(Circle(O,1));[/asy]$

$\textbf{(A)}\ x=3y\\ \textbf{(B)}\ x=2y\\ \textbf{(C)}\ x=60^\circ\\ \textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y\\ \textbf{(E)}\ x=2y\text{ or }x=3y\text{, depending upon the length of }AB$

Solution

Problem 45

Given a geometric sequence with the first term $\neq 0$ and $r \neq 0$ and an arithmetic sequence with the first term $=0$ . A third sequence $1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:

$\textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given}$

Solution

Problem 46

The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$ , and $y=\frac{2}{3}$ intersect in:

$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$

Solution

Problem 47

The expressions $a+bc$ and $(a+b)(a+c)$ are:

$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$

Solution

Problem 48

Given $\triangle ABC$ with medians $AE, BF, CD$ ; $FH$ parallel and equal to $AE$ ; $BH and HE$ are drawn; $FE$ extended meets $BH$ in $G$ . Which one of the following statements is not necessarily correct?

$\textbf{(A)}\ AEHF\text{ is a parallelogram}\qquad\textbf{(B)}\ HE=HG\\ \textbf{(C)}\ BH=DC\qquad\textbf{(D)}\ FG=\frac{3}{4}AB\qquad\textbf{(E)}\ FG\text{ is a median of triangle }BFH$

Solution

Problem 49

The graphs of $y=\frac{x^2-4}{x-2}$ and $y=2x$ intersect in:

$\textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points}$

Solution

Problem 50

In order to pass $B$ going $40$ mph on a two-lane highway, $A$ , going $50$ mph, must gain $30$ feet. Meantime, $C, 210$ feet from $A$ , is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, then, in order to pass safely, $A$ must increase his speed by:

$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$

Solution

@@ Line 1: / Line 1: @@
-== Problem 1 ==
+ == Problem 1==
+Which one of the following is not equivalent to <math>0.000000375</math>?
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\\ \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000} </math>
 [[1955 AHSME Problems/Problem 1|Solution]]
+== Problem 2==
-== Problem 2 ==
+The smaller angle between the hands of a clock at <math>12:25</math> p.m. is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ </math>
 [[1955 AHSME Problems/Problem 2|Solution]]
+== Problem 3==
-== Problem 3 ==
+If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the original ten numbers:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math>
 [[1955 AHSME Problems/Problem 3|Solution]]
+== Problem 4==
+The equality <math>\frac{1}{x-1}=\frac{2}{x-2}</math> is satisfied by:
-== Problem 4 ==
+<math> \textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0 </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
 [[1955 AHSME Problems/Problem 4|Solution]]
+== Problem 5==
-== Problem 5 ==
+<math>5y</math> varies inversely as the square of <math>x</math>. When <math>y=16, x=1</math>. When <math>x=8, y</math> equals:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024
 [[1955 AHSME Problems/Problem 5|Solution]]
+== Problem 6==
-== Problem 6 ==
+A merchant buys a number of oranges at <math>3</math> for <math>10</math> cents and an equal number at <math>5</math> for <math>20</math> cents. To "break even" he must sell all at:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents} </math>
 [[1955 AHSME Problems/Problem 6|Solution]]
+== Problem 7==
-== Problem 7 ==
+If a worker receives a <math>20</math>% cut in wages, he may regain his original pay exactly by obtaining a raise of:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{20%}\qquad\textbf{(B)}\ \text{25%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{%}\qquad\textbf{(D)}\ &#036;20\qquad\textbf{(E)}\ &#036;25 </math>
 [[1955 AHSME Problems/Problem 7|Solution]]
-== Problem 8 ==
+== Problem 8==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The graph of <math>x^2-4y^2=0</math>:
+<math> \textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does not exist} </math>
 [[1955 AHSME Problems/Problem 8|Solution]]
+== Problem 9==
-== Problem 9 ==
+A circle is inscribed in a triangle with sides <math>8, 15</math>, and <math>17</math>. The radius of the circle is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7   </math>
 [[1955 AHSME Problems/Problem 9|Solution]]
+== Problem 10==
-== Problem 10 ==
+How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each?
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40} </math>
 [[1955 AHSME Problems/Problem 10|Solution]]
+== Problem 11==
-== Problem 11 ==
+The negation of the statement "No slow learners attend this school" is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{All slow learners attend this school}\\ \textbf{(B)}\ \text{All slow learners do not attend this school}\\ \textbf{(C)}\ \text{Some slow learners attend this school}\\ \textbf{(D)}\ \text{Some slow learners do not attend this school}\\ \textbf{(E)}\ \text{No slow learners do not attend this school} </math>
 [[1955 AHSME Problems/Problem 11|Solution]]
+== Problem 12==
-== Problem 12 ==
+The solution of <math>\sqrt{5x-1}+\sqrt{x-1}=2</math> is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0 </math>
 [[1955 AHSME Problems/Problem 12|Solution]]
+== Problem 13==
-== Problem 13 ==
+The fraction <math>\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}</math> is equal to:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2 </math>
 [[1955 AHSME Problems/Problem 13|Solution]]
+== Problem 14==
-== Problem 14 ==
+The length of rectangle <math>R</math> is <math>10</math>% more than the side of square <math>S</math>. The width of the rectangle is <math>10</math>% less than the side of the square.
+The ratio of the areas, <math>R:S</math>, is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200 </math>
 [[1955 AHSME Problems/Problem 14|Solution]]
+== Problem 15==
-== Problem 15 ==
+The ratio of the areas of two concentric circles is <math>1: 3</math>. If the radius of the smaller is <math>r</math>, then the difference between the
+radii is best approximated by:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r    </math>
 [[1955 AHSME Problems/Problem 15|Solution]]
+== Problem 16==
-== Problem 16 ==
+The value of <math>\frac{3}{a+b}</math> when <math>a=4</math> and <math>b=-4</math> is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless} </math>
 [[1955 AHSME Problems/Problem 16|Solution]]
+== Problem 17==
-== Problem 17 ==
+If <math>\log x-5 \log 3=-2</math>, then <math>x</math> equals:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25} </math>
 [[1955 AHSME Problems/Problem 17|Solution]]
+== Problem 18==
-== Problem 18 ==
+The discriminant of the equation <math>x^2+2x\sqrt{3}+3=0</math> is zero. Hence, its roots are:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary} </math>
 [[1955 AHSME Problems/Problem 18|Solution]]
+== Problem 19==
-== Problem 19 ==
+Two numbers whose sum is <math>6</math> and the absolute value of whose difference is <math>8</math> are roots of the equation:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0 </math>
 [[1955 AHSME Problems/Problem 19|Solution]]
+== Problem 20==
-== Problem 20 ==
+The expression <math>\sqrt{25-t^2}+5</math> equals zero for:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5 </math>
 [[1955 AHSME Problems/Problem 20|Solution]]
+== Problem 21==
-== Problem 21 ==
+Represent the hypotenuse of a right triangle by <math>c</math> and the area by <math>A</math>. The altitude on the hypotenuse is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2} </math>
 [[1955 AHSME Problems/Problem 21|Solution]]
+== Problem 22==
-== Problem 22 ==
+On a \texdollar{10000} order a merchant has a choice between three successive discounts of <math>20</math>%, <math>20</math>%, and <math>10</math>% and
+three successive discounts of <math>40</math>%, <math>5</math>%, and <math>5</math>%. By choosing the better offer, he can save:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{nothing at all}\qquad\textbf{(B)}\ &#036;440\qquad\textbf{(C)}\ &#036;330\qquad\textbf{(D)}\ &#036;345\qquad\textbf{(E)}\ &#036;360 </math>
 [[1955 AHSME Problems/Problem 22|Solution]]
+== Problem 23==
-== Problem 23 ==
+In checking the petty cash a clerk counts <math>q</math> quarters, <math>d</math> dimes, <math>n</math> nickels, and <math>c</math> cents.
+Later he discovers that <math>x</math> of the nickels were counted as quarters and <math>x</math> of the dimes were counted as cents.
+To correct the total obtained the clerk must:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents} </math>
 [[1955 AHSME Problems/Problem 23|Solution]]
+== Problem 24==
-== Problem 24 ==
+The function <math>4x^2-12x-1</math>:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10} </math>
 [[1955 AHSME Problems/Problem 24|Solution]]
+== Problem 25==
-== Problem 25 ==
+One of the factors of <math>x^4+2x^2+9</math> is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1955 AHSME Problems/Problem 25|Solution]]
+== Problem 26==
-== Problem 26 ==
+Mr. A owns a house worth \textdollar{10000}. He sells it to Mr. <math>B</math> at <math>10</math>% profit. Mr. <math>B</math> sells the house back to Mr. <math>A</math> at a <math>10</math>% loss. Then:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{Mr. A comes out even}\qquad\textbf{(B)}\ \text{Mr. A makes }&#036;100\qquad\textbf{(C)}\ \text{Mr. A makes }&#036;1000\\ \textbf{(D)}\ \text{Mr. B loses }&#036;100\qquad\textbf{(E)}\ \text{none of the above is correct} </math>
 [[1955 AHSME Problems/Problem 26|Solution]]
+== Problem 27==
-== Problem 27 ==
+If <math>r</math> and <math>s</math> are the roots of <math>x^2-px+q=0</math>, then <math>r^2+s^2</math> equals:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2 </math>
 [[1955 AHSME Problems/Problem 27|Solution]]
+== Problem 28==
-== Problem 28 ==
+On the same set of axes are drawn the graph of <math>y=ax^2+bx+c</math> and the graph of the equation obtained by replacing <math>x</math> by <math>-x</math> in the given equation.
+If <math>b \neq 0</math> and <math>c \neq 0</math> these two graphs intersect:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis} </math>
 [[1955 AHSME Problems/Problem 28|Solution]]
+== Problem 29==
-== Problem 29 ==
+In the figure, <math>PA</math> is tangent to semicircle <math>SAR</math>; <math>PB</math> is tangent to semicircle <math>RBT</math>; <math>SRT</math> is a straight line;
+the arcs are indicated in the figure. <math>\angle APB</math> is measured by:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<asy>
+unitsize(1.2cm);
+defaultpen(linewidth(.8pt)+fontsize(8pt));
+dotfactor=3;
+pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0);
+pair A=O1+2*dir(60), B=O2+dir(85);
+pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2;
+pair P=extension(A,Pa,B,Pb);
+pair[] dots={Sp,R,T,A,B,P};
+draw(P--P+5*(A-P));
+draw(P--P+5*(B-P));
+clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle);
+draw(Arc(O1,2,0,180)--cycle);
+draw(Arc(O2,1,0,180)--cycle);
+dot(dots);
+label("$S$",Sp,S);
+label("$R$",R,S);
+label("$T$",T,S);
+label("$A$",A,NE);
+label("$B$",B,N);
+label("$P$",P,NNE);
+label("$a$",midpoint(Arc(O1,2,0,60)),SW);
+label("$b$",midpoint(Arc(O2,1,85,180)),SE);
+label("$c$",midpoint(Arc(O1,2,60,180)),SE);
+label("$d$",midpoint(Arc(O2,1,0,85)),SW);</asy>
+<math> \textbf{(A)}\ \frac{1}{2}(a-b)\qquad\textbf{(B)}\ \frac{1}{2}(a+b)\qquad\textbf{(C)}\ (c-a)-(d-b)\qquad\textbf{(D)}\ a-b\qquad\textbf{(E)}\ a+b </math>
 [[1955 AHSME Problems/Problem 29|Solution]]
+== Problem 30==
-== Problem 30 ==
+Each of the equations <math>3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}</math> has:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root} </math>
 [[1955 AHSME Problems/Problem 30|Solution]]
+== Problem 31==
-== Problem 31 ==
+An equilateral triangle whose side is <math>2</math> is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides.
+If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2} </math>
 [[1955 AHSME Problems/Problem 31|Solution]]
+== Problem 32==
-== Problem 32 ==
+If the discriminant of <math>ax^2+2bx+c=0</math> is zero, then another true statement about <math>a, b</math>, and <math>c</math> is that:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{they form an arithmetic progression}\\ \textbf{(B)}\ \text{they form a geometric progression}\\ \textbf{(C)}\ \text{they are unequal}\\ \textbf{(D)}\ \text{they are all negative numbers}\\ \textbf{(E)}\ \text{only b is negative and a and c are positive} </math>
 [[1955 AHSME Problems/Problem 32|Solution]]
+== Problem 33==
-== Problem 33 ==
+Henry starts a trip when the hands of the clock are together between <math>8</math> a.m. and <math>9</math> a.m.
+He arrives at his destination between <math>2</math> p.m. and <math>3</math> p.m. when the hands of the clock are exactly <math>180^\circ</math> apart. The trip takes:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. 43-7/11 min.}\qquad\textbf{(C)}\ \text{5 hr. 16-4/11 min.}\qquad\textbf{(D)}\ \text{6 hr. 30 min.}\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1955 AHSME Problems/Problem 33|Solution]]
+== Problem 34==
-== Problem 34 ==
+A <math>6</math>-inch and <math>18</math>-inch diameter pole are placed together and bound together with wire.
+The length of the shortest wire that will go around them is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi </math>
 [[1955 AHSME Problems/Problem 34|Solution]]
+== Problem 35==
-== Problem 35 ==
+Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles.
+The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy.
+The original number of marbles:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{is none of the following}\qquad\textbf{(B)}\ \text{cannot be determined from the given data}\\ \textbf{(C)}\ \text{is 20 or 26}\qquad\textbf{(D)}\ \text{is 14 or 32}\qquad\textbf{(E)}\ \text{is 8 or 38} </math>
 [[1955 AHSME Problems/Problem 35|Solution]]
+== Problem 36==
-== Problem 36 ==
+A cylindrical oil tank, lying horizontally, has an interior length of <math>10</math> feet and an interior diameter of <math>6</math> feet.
+If the rectangular surface of the oil has an area of <math>40</math> square feet, the depth of the oil is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 3-\sqrt{5}\qquad\textbf{(D)}\ 3+\sqrt{5}\\ \textbf{(E)}\ \text{either }3-\sqrt{5}\text{ or }3+\sqrt{5} </math>
 [[1955 AHSME Problems/Problem 36|Solution]]
+== Problem 37==
-== Problem 37 ==
+A three-digit number has, from left to right, the digits <math>h, t</math>, and <math>u</math>, with <math>h>u</math>.
+When the number with the digits reversed is subtracted from the original number, the units' digit in the difference of r.
+The next two digits, from right to left, are:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\qquad\textbf{(C)}\ \text{impossible to tell}\qquad\textbf{(D)}\ \text{5 and 4}\qquad\textbf{(E)}\ \text{4 and 5} </math>
 [[1955 AHSME Problems/Problem 37|Solution]]
+== Problem 38==
-== Problem 38 ==
+Four positive integers are given. Select any three of these integers, find their arithmetic average,
+and add this result to the fourth integer. Thus the numbers <math>29, 23, 21</math>, and <math>17</math> are obtained. One of the original integers is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17
 [[1955 AHSME Problems/Problem 38|Solution]]
+== Problem 39==
-== Problem 39 ==
+If <math>y=x^2+px+q</math>, then if the least possible value of <math>y</math> is zero <math>q</math> is equal to:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q </math>
 [[1955 AHSME Problems/Problem 39|Solution]]
+== Problem 40==
-== Problem 40 ==
+The fractions <math>\frac{ax+b}{cx+d}</math> and <math>\frac{b}{d}</math> are unequal if:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ a=c=1, x\neq 0\qquad\textbf{(B)}\ a=b=0\qquad\textbf{(C)}\ a=c=0\\ \textbf{(D)}\ x=0\qquad\textbf{(E)}\ ad=bc </math>
 [[1955 AHSME Problems/Problem 40|Solution]]
+== Problem 41==
-== Problem 41 ==
+A train traveling from Aytown to Beetown meets with an accident after <math>1</math> hr. It is stopped for <math>\frac{1}{2}</math> hr.,
+after which it proceeds at four-fifths of its usual rate, arriving at Beetown <math>2</math> hr. late.
+If the train had covered <math>80</math> miles more before the accident, it would have been just <math>1</math> hr. late.
+The usual rate of the train is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{20 mph}\qquad\textbf{(B)}\ \text{30 mph}\qquad\textbf{(C)}\ \text{40 mph}\qquad\textbf{(D)}\ \text{50 mph}\qquad\textbf{(E)}\ \text{60 mph} </math>
 [[1955 AHSME Problems/Problem 41|Solution]]
+== Problem 42==
-== Problem 42 ==
+If <math>a, b</math>, and <math>c</math> are positive integers, the radicals <math>\sqrt{a+\frac{b}{c}}</math> and <math>a\sqrt{\frac{b}{c}}</math> are equal when and only when:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{2}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text{ and }c=a-1 </math>
 [[1955 AHSME Problems/Problem 42|Solution]]
+== Problem 43==
-== Problem 43 ==
+The pairs of values of <math>x</math> and <math>y</math> that are the common solutions of the equations <math>y=(x+1)^2</math> and <math>xy+y=1</math> are:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{3 real pairs}\qquad\textbf{(B)}\ \text{4 real pairs}\qquad\textbf{(C)}\ \text{4 imaginary pairs}\\ \textbf{(D)}\ \text{2 real and 2 imaginary pairs}\qquad\textbf{(E)}\ \text{1 real and 2 imaginary pairs} </math>
 [[1955 AHSME Problems/Problem 43|Solution]]
+== Problem 44==
-== Problem 44 ==
+In circle <math>O</math> chord <math>AB</math> is produced so that <math>BC</math> equals a radius of the circle. <math>CO</math> is drawn and extended to <math>D</math>.
+<math>AO</math> is drawn. Which of the following expresses the relationship between <math>x</math> and <math>y</math>?
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ }  \qquad \textbf{(C) \ }  \qquad \textbf{(D) \ }  \qquad \textbf{(E) \ } </math>
+<asy>
+size(200);defaultpen(linewidth(0.7)+fontsize(10));
+pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0);
+draw(O--A--C--D);
+dot(A^^B^^C^^D^^O);
+pair point=O;
+label("$A$", A, dir(point--A));
+label("$B$", B, dir(point--B));
+label("$C$", C, dir(point--C));
+label("$D$", D, dir(point--D));
+label("$O$", O, dir(285));
+label("$x$", O+0.1*dir(172.5), dir(172.5));
+label("$y$", C+0.4*dir(187.5), dir(187.5));
+draw(Circle(O,1));</asy>
+<math> \textbf{(A)}\ x=3y\\ \textbf{(B)}\ x=2y\\ \textbf{(C)}\ x=60^\circ\\ \textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y\\ \textbf{(E)}\ x=2y\text{ or }x=3y\text{, depending upon the length of }AB </math>
 [[1955 AHSME Problems/Problem 44|Solution]]
+== Problem 45==
-== Problem 45 ==
+Given a geometric sequence with the first term <math>\neq 0</math> and <math>r \neq 0</math> and an arithmetic sequence with the first term <math>=0</math>.
+A third sequence <math>1,1,2\ldots</math> is formed by adding corresponding terms of the two given sequences.
+The sum of the first ten terms of the third sequence is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given} </math>
 [[1955 AHSME Problems/Problem 45|Solution]]
+== Problem 46==
-== Problem 46 ==
+The graphs of <math>2x+3y-6=0, 4x-3y-6=0, x=2</math>, and <math>y=\frac{2}{3}</math> intersect in:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points} </math>
 [[1955 AHSME Problems/Problem 46|Solution]]
+== Problem 47==
-== Problem 47 ==
+The expressions <math>a+bc</math> and <math>(a+b)(a+c)</math> are:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0 </math>
 [[1955 AHSME Problems/Problem 47|Solution]]
+== Problem 48==
-== Problem 48 ==
+Given <math>\triangle ABC</math> with medians <math>AE, BF, CD</math>; <math>FH</math> parallel and equal to <math>AE</math>; <math>BH and HE</math> are drawn; <math>FE</math> extended meets <math>BH</math> in <math>G</math>.
+Which one of the following statements is not necessarily correct?
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ AEHF\text{ is a parallelogram}\qquad\textbf{(B)}\ HE=HG\\ \textbf{(C)}\ BH=DC\qquad\textbf{(D)}\ FG=\frac{3}{4}AB\qquad\textbf{(E)}\ FG\text{ is a median of triangle }BFH </math>
 [[1955 AHSME Problems/Problem 48|Solution]]
+== Problem 49==
-== Problem 49 ==
+The graphs of <math>y=\frac{x^2-4}{x-2}</math> and <math>y=2x</math> intersect in:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points} </math>
 [[1955 AHSME Problems/Problem 49|Solution]]
+== Problem 50==
-== Problem 50 ==
+In order to pass <math>B</math> going <math>40</math> mph on a two-lane highway, <math>A</math>, going <math>50</math> mph, must gain <math>30</math> feet.
+Meantime, <math>C, 210</math> feet from <math>A</math>, is headed toward him at <math>50</math> mph. If <math>B</math> and <math>C</math> maintain their speeds,
+then, in order to pass safely, <math>A</math> must increase his speed by:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph} </math>
+[[1955 AHSME Problems/Problem 50|Solution]]
-[[1955 AHSME Problems/Problem 50|Solution]]
 == See also ==
 * [[AHSME]]
 * [[AHSME Problems and Solutions]]
+* [[1950 AHSME]]
 * [[Mathematics competition resources]]
 {{MAA Notice}}

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Difference between revisions of "1955 AHSME Problems"

Revision as of 08:43, 13 October 2014

Contents

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

See also